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ConcurrentHashMap笔记
概览:
内部存储的数据结构为:数组+链表+红黑树,图示:
重要的属性(内部类):
//存放元素的数组 transient volatile Node<K,V>[] table; //数组中的Node节点 static class Node<K,V> implements Map.Entry<K,V> { final int hash;//Key计算出来的Hash值 final K key;//Key volatile V val;//Value volatile Node<K,V> next;//链表的下一个节点 Node(int hash, K key, V val, Node<K,V> next) { this.hash = hash; this.key = key; this.val = val; this.next = next; } ...//省略 } //红黑树中的节点 static final class TreeNode<K,V> extends Node<K,V> { TreeNode<K,V> parent; // red-black tree links TreeNode<K,V> left; //左子节点 TreeNode<K,V> right; //右子节点 TreeNode<K,V> prev; // boolean red; ...//省略 } //组合TreeNode static final class TreeBin<K,V> extends Node<K,V> { TreeNode<K,V> root; volatile TreeNode<K,V> first; } //内部类
方法分析
/** Implementation for put and putIfAbsent */ final V putVal(K key, V value, boolean onlyIfAbsent) { if (key == null || value =http://www.mamicode.com/= null) throw new NullPointerException(); int hash = spread(key.hashCode()); int binCount = 0; for (Node<K,V>[] tab = table;;) { Node<K,V> f; int n, i, fh; if (tab == null || (n = tab.length) == 0) tab = initTable();//初始化数组大小,默认16 //数组指定位置元素为空,直接插入 else if ((f = tabAt(tab, i = (n - 1) & hash)) == null) { if (casTabAt(tab, i, null, new Node<K,V>(hash, key, value, null))) break; // no lock when adding to empty bin } else if ((fh = f.hash) == MOVED) tab = helpTransfer(tab, f); else {//不为空,链表存储 V oldVal = null; synchronized (f) { if (tabAt(tab, i) == f) { if (fh >= 0) { binCount = 1; for (Node<K,V> e = f;; ++binCount) { K ek; if (e.hash == hash && ((ek = e.key) == key || (ek != null && key.equals(ek)))) { oldVal = e.val; if (!onlyIfAbsent) e.val = value; break; } Node<K,V> pred = e; if ((e = e.next) == null) { pred.next = new Node<K,V>(hash, key, value, null); break; } } } //红黑树 else if (f instanceof TreeBin) { Node<K,V> p; binCount = 2; if ((p = ((TreeBin<K,V>)f).putTreeVal(hash, key, value)) != null) { oldVal = p.val; if (!onlyIfAbsent) p.val = value; } } } } if (binCount != 0) { //链表长度大于8,转为红黑树存储 if (binCount >= TREEIFY_THRESHOLD) treeifyBin(tab, i); if (oldVal != null) return oldVal; break; } } } addCount(1L, binCount); return null; }
/** * Replaces all linked nodes in bin at given index unless table is * too small, in which case resizes instead. */ private final void treeifyBin(Node<K,V>[] tab, int index) { Node<K,V> b; int n, sc; if (tab != null) { if ((n = tab.length) < MIN_TREEIFY_CAPACITY) tryPresize(n << 1);//扩容数组 else if ((b = tabAt(tab, index)) != null && b.hash >= 0) { synchronized (b) { if (tabAt(tab, index) == b) { TreeNode<K,V> hd = null, tl = null;
//遍历链表,构造TreeNode for (Node<K,V> e = b; e != null; e = e.next) { TreeNode<K,V> p = new TreeNode<K,V>(e.hash, e.key, e.val, null, null); if ((p.prev = tl) == null) hd = p; else tl.next = p; tl = p; }
//构建红黑树 setTabAt(tab, index, new TreeBin<K,V>(hd)); } } } } }
public V get(Object key) { Node<K,V>[] tab; Node<K,V> e, p; int n, eh; K ek; int h = spread(key.hashCode()); if ((tab = table) != null && (n = tab.length) > 0 && (e = tabAt(tab, (n - 1) & h)) != null) { if ((eh = e.hash) == h) { if ((ek = e.key) == key || (ek != null && key.equals(ek))) return e.val; } else if (eh < 0)//红黑树取值 return (p = e.find(h, key)) != null ? p.val : null;
//链表取值 while ((e = e.next) != null) { if (e.hash == h && ((ek = e.key) == key || (ek != null && key.equals(ek)))) return e.val; } } return null; }
写在最后:
为什么使用红黑树?
红黑树的特性:
1、节点是红色或者黑色
2、根是黑色
3、所有叶子都是黑色
4、每个红色节点必须有2个黑色的子节点
5、从任一节点到其每个叶子的所有简单路径包含相同数目的黑色节点
根据特性5,从根的最长路径不可能>2倍的最短路径,所以这样的二叉树是平衡的;插入、删除、查询操作比较高效
拓展阅读
1、红黑树介绍 2、深入分析ConcurrentHashMap1.8的扩容实现
ConcurrentHashMap笔记
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